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Question

Question: How do you solve \(2{\sin ^2}x = 1\)?...

How do you solve 2sin2x=12{\sin ^2}x = 1?

Explanation

Solution

This is a trigonometric problem and it can be solved with basic properties of trigonometry. Here, we have to find the value of x. For which we need to simplify the equation and bring it to the form of sin x. And then solve it by placing the sin value.

Complete step by step solution:
The given equation is 2sin2x=12{\sin ^2}x = 1
We will take the co-efficient of sin2x{\sin ^2}x i.e., 22 on the other side of equal to
sin2x=12\therefore {\sin ^2}x = \dfrac{1}{2}
Now, we will take the square root of both the side
sinx=±12\Rightarrow \sin x = \pm \dfrac{1}{{\sqrt 2 }}
Here first we consider sinx=12\sin x = \dfrac{1}{{\sqrt 2 }}
Now put sin45=12\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}, we get,
sinx=sinπ4\Rightarrow \sin x = \sin \dfrac{\pi }{4}
Since 45=π4{45^ \circ } = \dfrac{\pi }{4}in radian.
Similarly, now considersinx=12\sin x = - \dfrac{1}{{\sqrt 2 }}
Now put sin45=12\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}, but as it is minus it will be sin(π4)\sin ( - \dfrac{\pi }{4}), we get,
sinx=sin(π4)\Rightarrow \sin x = \sin ( - \dfrac{\pi }{4})
When we solve the above equation
sinx=sin(ππ4)\Rightarrow \sin x = \sin (\pi - \dfrac{\pi }{4})
sinx=sin(4ππ4)\Rightarrow \sin x = \sin (\dfrac{{4\pi - \pi }}{4})
So, after simplifying we get,
sinx=sin3π4\Rightarrow \sin x = \sin \dfrac{{3\pi }}{4}
So we will consider all the values in all the quadrants of sinx\sin x where x=π4,3π4,5π4,7π4,.............x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{5\pi }}{4},\dfrac{{7\pi }}{4},.............
sinx=sin(π4+π2n)\therefore \sin x = \sin (\dfrac{\pi }{4} + \dfrac{\pi }{2}n)
where, n is 1,2,3,4,…… or we can say nRn \in R
x=π4+π2n\therefore x = \dfrac{\pi }{4} + \dfrac{\pi }{2}n.

Additional Information:
The sinx\sin x function gives progressive values that mean from sin0\sin {0^ \circ } to sin90\sin {90^ \circ } the value of sinx\sin x will increase. So, some basic values of sinθ\sin \theta where θ\theta is 0,30,45,60{0^ \circ },{30^ \circ },{45^ \circ },{60^ \circ }and 90{90^ \circ }.
sin0=0,sin30=12,sin45=12,sin60=32,sin90=1\therefore sin{0^ \circ } = 0,\sin {30^ \circ } = \dfrac{1}{2},\sin {45^ \circ } = \sqrt {\dfrac{1}{2}} ,\sin {60^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {90^ \circ } = 1.

Note: In order to solve this problem, we must learn about the values of sinθ\sin \theta where θ\theta can be any degree. We also have to know about the values of sin30,sin45\sin {30^ \circ },\sin {45^ \circ } and sin90\sin {90^ \circ }. The trigonometric function sinx\sin x range lies [1,1][ - 1,1] and it is a closed interval that means it includes -1 as well as 1. All the real numbers are the domain of the trigonometric function sinx\sin x .
The sign convention should be taken care of because the value of sin x has a square root in it. And we already know that the square of a positive or negative number is always positive.
We should always keep in mind that the square root has plus/minus signs which state that we require both the positive integer as well as negative integers.